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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

4. Rational Expressions

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Exercise 62 Page 533

What is the inverse operation of a square root?

x=168

Practice makes perfect

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

sqrt(x+1)-5 = 8

LHS+5=RHS+5

sqrt(x+1) = 13

LHS^2=RHS^2

(sqrt(x+1))^2=13^2

( sqrt(a) )^2 = a

x+1 = 13^2

Calculate power

x+1 = 169

LHS-1=RHS-1

x = 168

Finally, we will check our solution by substituting the value into the original equation.

sqrt(x+1)-5 = 8

x= 168

sqrt(168+1)-5 ? = 8

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Simplify left-hand side

Add terms

sqrt(169)-5 ? = 8

Calculate root

13-5 ? = 8

Subtract term

8=8 ✓

Subchapter links

Lesson Check p.530

Standardized Test Prep p.533

Mixed Review p.533